Prove that $\mathbb{P}(T_y < \infty) = \frac{a}{y}$

Question

Problem :

Let $M$ be a continuous non negative martingale such that $M_{0}=a>0$ and $\lim _{t \rightarrow \infty} M_{t}=0$ a.s.

1. For $y \geq a,$ let $T_{y}=\inf \left\{t \geq 0, M_{t}=y\right\} .$ Prove that $\mathbb{P}\left(T_{y}<\infty\right)=a / y$
2. Prove that $\sup _{t \geq 0} M_{t} \sim \frac{a}{U}$ where $U \sim \mathcal{U}([0,1])$

My attempt :

both $T_y \wedge n $ and $0$ are bounded stopping times, according to Doob's optional stopping theorem, we have :

$$\mathbb{E}(M_{T_y \wedge n}) = \mathbb{E}(M_0) = a$$

on the other hand we have :

\begin{align*} \mathbb{E}(M_{T_y \wedge n}) &= \mathbb{E}(M_{T_y \wedge n} | T_y < \infty )P(T_y < \infty) + \mathbb{E}(M_{T_y \wedge n} | T_y = \infty )P(T_y = \infty) \\ & = \mathbb{E}(M_{T_y \wedge n} )P(T_y < \infty) + \mathbb{E}(M_n)P(T_y = \infty) \\ \end{align*}

since $M_{T_y \wedge n} \to M_{T_y}$ in $L^1$ then if $M_n \to 0$ in $L^1$ then question 1. is proven.

but do we have $M_n \to 0$ in $L^1$ ?

I know that a necessary condition to obtain the above is that the martingale is uniformly integrable, but in this problem it doesn't look like it's uniformly integrable.

am I tackling the problem the wrong way ?

Answer 1

As you wrote we have: $$a = \mathbb{E}(M_{T_y \wedge n} | T_y < \infty )P(T_y < \infty) + \mathbb{E}(M_{T_y \wedge n} | T_y = \infty )P(T_y = \infty)$$

Now the martingale $M_{T_y \wedge n}$ on the event $\{T_y = \infty\}$ is bounded and $\lim_n M_{T_y \wedge n}=0$. Hence, by dominated convergence $\lim_n\mathbb{E}(M_{T_y \wedge n} | T_y = \infty )=0$.

Also, again $M_{T_y \wedge n}$ on the event $\{T_y < \infty\}$ is bounded and $\lim_n M_{T_y \wedge n}=y$. Hence, by dominated convergence $\lim_n\mathbb{E}(M_{T_y \wedge n} | T_y < \infty )=y$.

Remark: The equalities $\mathbb{E}(M_{T_y \wedge n} | T_y = \infty )= \mathbb{E}(M_{n} )$ and $\mathbb{E}(M_{T_y \wedge n} | T_y < \infty )= \mathbb{E}(M_{T_y \wedge n} )$ are not true in general.